我想知道如何在基于X，Y鼠标坐标的android上执行拖动吗？
考虑两个简单的示例，Team Viewer / QuickSupport 分别
在远程智能手机和Windows Paint Pen上绘制“密码模式”
。

我所能做的就是模拟触摸（使用dispatchGesture()和也AccessibilityNodeInfo.ACTION_CLICK）。

我找到了这些相关链接，但不知道它们是否有用：

使用AccessibilityService在屏幕上执行滑动例子1
继续手势下面是我的工作代码，该代码用于将鼠标坐标（在PictureBox控件内部）发送到远程电话并模拟触摸。

Windows窗体应用程序：

private void pictureBox1_MouseDown(object sender, MouseEventArgs e)
{
    foreach (ListViewItem item in lvConnections.SelectedItems)
    {
        // Remote screen resolution
        string[] tokens = item.SubItems[5].Text.Split('x'); // Ex: 1080x1920

        int xClick = (e.X * int.Parse(tokens[0].ToString())) / (pictureBox1.Size.Width);
        int yClick = (e.Y * int.Parse(tokens[1].ToString())) / (pictureBox1.Size.Height);

        Client client = (Client)item.Tag;

        if (e.Button == MouseButtons.Left)
            client.sock.Send(Encoding.UTF8.GetBytes("TOUCH" + xClick + "<|>" + yClick + Environment.NewLine));
    }
}

编辑：

我最后的尝试是分别使用鼠标坐标（“ C＃Windows窗体
应用程序”）和自定义android例程（参考
上面链接的“滑动屏幕”的代码）进行“滑动屏幕”：

private Point mdownPoint = new Point();

private void pictureBox1_MouseDown(object sender, MouseEventArgs e)
{
    foreach (ListViewItem item in lvConnections.SelectedItems)
    {
        // Remote screen resolution
        string[] tokens = item.SubItems[5].Text.Split('x'); // Ex: 1080x1920

        Client client = (Client)item.Tag;

        if (e.Button == MouseButtons.Left)
        {
            xClick = (e.X * int.Parse(tokens[0].ToString())) / (pictureBox1.Size.Width); 
            yClick = (e.Y * int.Parse(tokens[1].ToString())) / (pictureBox1.Size.Height);

            // Saving start position:

            mdownPoint.X = xClick; 
            mdownPoint.Y = yClick;

            client.sock.Send(Encoding.UTF8.GetBytes("TOUCH" + xClick + "<|>" + yClick + Environment.NewLine));
        }
    }
}

private void PictureBox1_MouseMove(object sender, MouseEventArgs e)
{
    foreach (ListViewItem item in lvConnections.SelectedItems)
    {
        // Remote screen resolution
        string[] tokens = item.SubItems[5].Text.Split('x'); // Ex: 1080x1920

        Client client = (Client)item.Tag;

        if (e.Button == MouseButtons.Left)
        {
            xClick = (e.X * int.Parse(tokens[0].ToString())) / (pictureBox1.Size.Width);
            yClick = (e.Y * int.Parse(tokens[1].ToString())) / (pictureBox1.Size.Height);

            client.sock.Send(Encoding.UTF8.GetBytes("MOUSESWIPESCREEN" + mdownPoint.X + "<|>" + mdownPoint.Y + "<|>" + xClick + "<|>" + yClick + Environment.NewLine));
        }
    }
}

android AccessibilityService :

public void Swipe(int x1, int y1, int x2, int y2, int time) {

if (android.os.Build.VERSION.SDK_INT >= android.os.Build.VERSION_CODES.N) {
    System.out.println(" ======= Swipe =======");

    GestureDescription.Builder gestureBuilder = new GestureDescription.Builder();
    Path path = new Path();
    path.moveTo(x1, y1);
    path.lineTo(x2, y2);

    gestureBuilder.addStroke(new GestureDescription.StrokeDescription(path, 100, time));
    dispatchGesture(gestureBuilder.build(), new GestureResultCallback() {
        @Override
        public void onCompleted(GestureDescription gestureDescription) {
            System.out.println("SWIPE Gesture Completed :D");
            super.onCompleted(gestureDescription);
        }
    }, null);
}

}

会产生以下结果（例如，仍然无法绘制“图案
密码”，例如TeamViewer）。但是，就像下面的评论中所说的那样，我
认为使用类似的方法可以使用“继续
手势”来实现。
欢迎向这个方向提出任何建议。

Edit 2:

Definitely, the solution is continued
gestures
like said on previous Edit.

• Simulating joystick movement using AccessibilityService
• Why the continueStroke function is not work

And below is a supposed fixed code that i found
here =>

android AccessibilityService:

// Simulates an L-shaped drag path: 200 pixels right, then 200 pixels down.
Path path = new Path();
path.moveTo(200,200);
path.lineTo(400,200);

final GestureDescription.StrokeDescription sd = new GestureDescription.StrokeDescription(path, 0, 500, true);

// The starting point of the second path must match
// the ending point of the first path.
Path path2 = new Path();
path2.moveTo(400,200);
path2.lineTo(400,400);

final GestureDescription.StrokeDescription sd2 = sd.continueStroke(path2, 0, 500, false); // 0.5 second

HongBaoService.mService.dispatchGesture(new GestureDescription.Builder().addStroke(sd).build(), new AccessibilityService.GestureResultCallback(){

@Override
public void onCompleted(GestureDescription gestureDescription){
super.onCompleted(gestureDescription);
HongBaoService.mService.dispatchGesture(new GestureDescription.Builder().addStroke(sd2).build(),null,null);
}

@Override
public void onCancelled(GestureDescription gestureDescription){
super.onCancelled(gestureDescription);
}
},null);

Then, my doubt is: how send correctly mouse coordinates for code above, of
the way that can perform drag to any direction? Some idea?

然后，我的疑问是：如何正确发送上面代码的鼠标坐标，
以哪种方式可以向任意方向执行拖动？有想法吗

编辑3：

我发现了两个用于执行拖动的例程，但是它们使用的是
UiAutomation
+ injectInputEvent()。AFAIK，
事件注入仅在像
此处和
此处所述的系统应用程序中有效，我不希望
如此。

这是发现的例程：

公共布尔刷卡（int downX，int downY，int upX，int upY，int步骤，布尔拖动）
公共布尔刷卡（Point []段，int segmentSteps）
然后实现我的目标，我认为2RD例行更拨付给使用
具有表现出对码（逻辑下，不包括事件注）编辑
2和发送的所有点pictureBox1_MouseDown和pictureBox1_MouseMove
（C＃Windows窗体应用程序）分别填写Point[]动态，并
在pictureBox1_MouseUp发送cmd执行例程并使用此数组
填充。如果您对第一个例程有想法，请告诉我：D。

如果在阅读完此编辑后您有可能的解决方案，
请在回答中告诉我，而我将尝试测试该想法。